Systems of Inequalities

Intuition

The last two lessons treated a system’s solution set as an object in its own right, and for linear equations that object was always thin. It was a single point where the lines cross, a whole line when they coincide, or nothing at all when the system is inconsistent. Replace even one equals sign with an inequality and the picture changes completely. A single point has no room inside it, but a limit describes a whole range of allowable choices. Such a limit might be “you have at most $10$ hours” or “spend no more than your budget”. The solution set swells from a point into a two-dimensional region. This lesson is about those regions: how one inequality splits the plane, and how several of them together carve out the shape you are allowed to live in. The lesson then shows how to pick the single best point out of the infinitely many the region holds.

Learning goals

Shade the half-plane a test point satisfies
Draw the boundary solid or dashed by the symbol
Intersect the half-planes to get the feasible region
Find the corners by solving the boundary-line systems
Evaluate a linear objective at every corner
Allow for an empty or unbounded region

A single inequality carves the plane in two

Start with one linear inequality in two variables, such as $x + 2y < 6$ . Its solutions are all the pairs $(x, y)$ that make it true, and there are infinitely many of them. To see their shape, first draw the boundary line $x + 2y = 6$ , the set where the two sides are exactly equal. That line does not solve the inequality (equality is not “less than”), but it is the fence between the pairs that do and the pairs that do not.

The key fact is that the line splits the plane into two sides, and each side is entirely solutions or entirely non-solutions, with no mixing. So you never test more than one point: check a single point off the line, and its verdict settles the whole side it belongs to. Here is why that works.

Why the solutions of $ax + by < c$ fill one whole side of the line#

The boundary is the line $ax + by = c$ . Take the common case $b > 0$ first, and solve the boundary for $y$ :

y = \frac{c - ax}{b}.

At a given $x$ , this is the height of the line. A point $(x, y)$ sits above the line exactly when its own height beats the line’s height there, $y > \frac{c - ax}{b}$ . Multiply both sides by the positive number $b$ ; from the lesson on linear inequalities, a positive multiplier keeps an inequality’s direction, so

by > c - ax, \quad\text{that is}\quad ax + by > c.

So “above the line” is exactly the set where $ax + by > c$ , and running the same steps with the inequality reversed shows “below the line” is exactly where $ax + by < c$ . Every point of the plane is above, on, or below the line, so these three conditions partition the plane, and each strict inequality names one full side.

If instead $b < 0$ , multiplying by $b$ reverses the direction (again from the linear inequalities lesson), so “above” becomes $ax + by < c$ : the side attached to each symbol flips. That is precisely why you cannot read the shaded side straight off the symbol without watching the sign of $b$ . That direction flip is also why the safe method is to test an actual point. And if $b = 0$ the boundary is the vertical line $x = \frac{c}{a}$ , which splits the plane into $x < \frac{c}{a}$ and $x > \frac{c}{a}$ ; there you solve for $x$ instead of $y$ .

In every case the boundary cuts the plane into two sides, one where $ax + by > c$ and one where $ax + by < c$ . Testing a single point off the line reveals which side is which.

That gives the whole procedure for graphing one inequality. Draw the boundary line. Make it solid when the inequality allows equality ( $\le$ or $\ge$ ), because then the line’s own points are solutions. Make it dashed when the inequality does not ( $<$ or $>$ ), because then those points are excluded. Then pick a test point not on the line and substitute it. Shade the side the test point lands on if it makes the inequality true, or the other side if it makes it false. The origin $(0, 0)$ is the easiest test point whenever the line does not pass through it.

Worked example 1 Graph the inequality $x + 2y < 6$

Begin with the boundary line $x + 2y = 6$ . Two points fix it. Setting $x = 0$ gives $2y = 6$ , so $y = 3$ and the point $(0, 3)$ ; setting $y = 0$ gives $x = 6$ , the point $(6, 0)$ . The inequality is strict, so the boundary is drawn dashed: points on the line are not solutions.

Now choose which side to shade by testing one point off the line. The origin is not on this line, so use it:

0 + 2(0) = 0, \quad 0 < 6 \text{ is true}.

Because $(0, 0)$ satisfies the inequality, the whole side of the line containing the origin is solutions. Shade that side. The solution set is that entire open half-plane, with the dashed boundary excluded.

The boundary line splits the plane. Testing the origin (a true statement here) shows the solutions fill the lower-left side; the dashed line marks a strict inequality, so the boundary itself is excluded.

Check your understanding

Is the point $(0,0)$ a solution of $3x - y \ge 6$ , and should its boundary line be drawn solid or dashed?

A Yes; dashed B No; solid C Yes; solid D No; dashed

Stacking inequalities: the feasible region

A system of inequalities asks for all of its conditions at once. A pair $(x, y)$ solves the system only when it solves every inequality, so it must lie in every one of the half-planes at the same time. The solution set is therefore the intersection of the half-planes: the region where all the shaded sides overlap. This is the two-variable echo of the “and” from one-variable inequalities, where joining conditions with “and” meant intersecting their solution sets. The overlap region has a name worth knowing, the feasible region, because in an applied problem it is exactly the set of choices that are actually allowed.

Here is a concrete system to graph. A workshop builds $x$ tables and $y$ bookshelves. Cutting a table takes $2$ hours and a bookshelf $1$ hour, with $10$ cutting hours available, so $2x + y \le 10$ . Assembly takes $1$ hour for either piece, with $7$ assembly hours available, so $x + y \le 7$ . You cannot build a negative count of anything, which forces $x \ge 0$ and $y \ge 0$ . Those four inequalities are the system, and its solution set is every build plan that respects all of the limits together.

Worked example 2 Graph the system and find its corners

The four inequalities are

2x + y \le 10, \qquad x + y \le 7, \qquad x \ge 0, \qquad y \ge 0.

The two conditions $x \ge 0$ and $y \ge 0$ confine everything to the first quadrant. Within it, graph each slanted boundary and keep the side each inequality allows. Here $2x + y \le 10$ holds below its line (test the origin: $0 \le 10$ is true), and $x + y \le 7$ holds below its line as well. Keeping only the common overlap leaves the four-sided feasible region shown below.

Its corners are where two boundaries cross, and each corner is the solution of a two-equation system. The interesting corner is where cutting and assembly are both used to the full, where $2x + y = 10$ and $x + y = 7$ meet. Subtract the second equation from the first to eliminate $y$ :

(2x + y) - (x + y) = 10 - 7 \;\Rightarrow\; x = 3.

Then $y = 7 - 3 = 4$ , giving the corner $(3, 4)$ . The other three corners are read off directly: the origin $(0, 0)$ , the point $(5, 0)$ where $2x + y = 10$ meets the $x$ -axis, and $(0, 7)$ where $x + y = 7$ meets the $y$ -axis. So the feasible region is the quadrilateral with corners $(0, 0)$ , $(5, 0)$ , $(3, 4)$ , and $(0, 7)$ .

The feasible region is the overlap of all four half-planes: the shaded quadrilateral. Each corner is where two boundary lines cross, found by solving that pair of equations.

The same test runs in reverse, which is how you write a system from a picture. Given a shaded region, each boundary line supplies one inequality: write the line’s equation, then fix the direction by testing a point you know is inside the region. That interior point must satisfy the inequality, so choose the symbol that makes it true. Use $\le$ or $\ge$ when the boundary is solid, and $<$ or $>$ when it is dashed. Doing this for every boundary rebuilds the whole system from the drawing.

Check your understanding

The two boundary lines $x + y = 5$ and $2x - y = 4$ meet at a corner of a feasible region. What are its coordinates?

(3, 2)

(2, 3)

(1, 4)

(4, 1)

Optimizing over the region: linear programming

A feasible region collects every allowed choice, but usually you want the best one. In the workshop, “best” means the most profit. Suppose each table earns a profit of $30$ dollars and each bookshelf $20$ dollars. Measuring profit in tens of dollars, the workshop wants to maximize the objective

P = 3x + 2y

over the feasible region. Maximizing or minimizing a linear objective over a region cut out by linear inequalities is called linear programming. When the region is a bounded polygon with corners, linear programming has a strikingly simple answer, that you need only check the corners.

Why a linear objective is optimized at a corner#

Fix the objective $P = 3x + 2y$ . For each number $c$ , the equation $3x + 2y = c$ is a straight line, and because they all share the slope $-\tfrac{3}{2}$ , the lines for different values of $c$ are parallel. They form a family sweeping across the plane, and moving to a larger $c$ shifts the line toward the upper right, the direction in which $P$ grows.

Maximizing $P$ over the feasible region means finding the largest $c$ for which the line $3x + 2y = c$ still touches the region. Picture sliding the line in the increasing direction until it is about to leave: the last point it touches is where $P$ is greatest on the region.

The feasible region is convex, meaning the straight segment joining any two of its points stays entirely inside it. This holds because each inequality’s half-plane is convex, and the region is their intersection, so a segment lying inside every half-plane lies inside the region. On a convex polygon, the final contact of a sliding line is either a corner or an edge parallel to that line. In the edge case, its endpoints are corners. Either way a corner attains the maximum.

The same argument with the line sliding the other way puts a minimizer at a corner too, which gives the corner point principle. Stated carefully, the principle applies to a feasible region whose boundaries are solid and which has at least one corner. On such a region, wherever an optimum of a linear objective exists, at least one corner attains it. So to optimize over a bounded polygon of that kind, evaluate the objective at each corner and take the best value. The qualifiers earn their place. If the region is unbounded in the increasing direction, the slide never stops and no maximum exists. If a boundary is dashed, so the region is open, the objective can creep toward a best value it never actually reaches. And when the objective line is parallel to an edge, that whole edge is optimal, though its corners still attain the value. Whenever an optimum exists and the region has corners, a corner is among the points that attain it.

The objective lines P = 3x + 2y are parallel; larger P sits farther up and to the right. Sliding one across the region, the last corner it touches is (3, 4), so the maximum profit is there.

Worked example 3 Find the most profitable plan

The feasible region from the previous example has corners $(0, 0)$ , $(5, 0)$ , $(3, 4)$ , and $(0, 7)$ . By the corner point principle, the maximum of $P = 3x + 2y$ sits at one of them, so evaluate $P$ at all four:

P(0,0) = 0, \quad P(5,0) = 15, \quad P(0,7) = 14, \quad P(3,4) = 17.

The largest value is $17$ , at the corner $(3, 4)$ . So the best plan builds $3$ tables and $4$ bookshelves, using all $10$ cutting hours and all $7$ assembly hours. That plan earns a profit of $17$ tens of dollars, that is $170$ dollars.

Notice that neither lopsided plan wins. Building as many tables as possible, $(5, 0)$ , earns $15$ , and leaning hard on bookshelves, $(0, 7)$ , earns $14$ . The optimum mixes the two products, which is exactly the kind of balance linear programming is built to find.

Check your understanding

A bounded feasible region has corners $(0, 6)$ , $(2, 2)$ , and $(6, 0)$ . What is the minimum of $C = 4x + 3y$ on the region?

18

14

24

8

When the region is empty or unbounded

Not every system produces a tidy polygon. Two other outcomes carry straight over from the consistent-or-inconsistent vocabulary of the equation lessons, and both matter for optimization.

Worked example 4 A region can be empty or unbounded

Two quick cases show the range of possibilities.

First, an empty region. The system $x + y \ge 5$ and $x + y \le 2$ asks for a sum that is at least $5$ and at most $2$ at the same time. No point can satisfy both, since a number cannot be $\ge 5$ and $\le 2$ at once, so the two half-planes never overlap and the feasible region is empty, $\varnothing$ . This is the two-variable version of an inconsistent system: the constraints contradict one another and no choice can meet them all. There is nothing to optimize.

Second, an unbounded region. The system $x \ge 0$ , $y \ge 0$ , and $x + y \ge 4$ keeps everything in the first quadrant that lies on or above the line $x + y = 4$ , and that region runs outward forever toward the upper right. On it the objective $x + y$ can be made as large as you like. So the objective has no maximum, but it does have a minimum of $4$ , reached all along the edge $x + y = 4$ , including the corners $(4, 0)$ and $(0, 4)$ . So on an unbounded region an optimum can fail to exist in one direction while still existing in the other. Whenever an optimum does exist on such a region, at least one corner attains it.

Common mistakes

Shading the side without testing. The symbol alone does not tell you the side once the inequality is not solved for $y$ . Substitute a test point (the origin is easiest) and shade the side that makes the statement true. Never guess the side from the direction of the symbol.
Testing a point that lies on the boundary. A test point on the line gives equality, which decides nothing. Pick a point clearly off the line; if the boundary passes through the origin, test something like $(1, 0)$ instead.
Mixing up solid and dashed. A solid boundary means the line’s points are included ( $\le$ or $\ge$ ); a dashed boundary means they are excluded ( $<$ or $>$ ). The corners of a feasible region only count as solutions when their boundaries are solid.
Forgetting the hidden constraints. In a word problem you almost always cannot use a negative amount, so $x \ge 0$ and $y \ge 0$ are part of the system even when the story never says so. Leaving them out can add phantom corners with negative coordinates.
Dividing by a negative without reversing the symbol. If you rewrite an inequality by multiplying or dividing by a negative to solve for $y$ , the symbol must flip. Keep the old direction and you will shade the wrong side. This is the same negative-flip rule from the lesson on linear inequalities. A correct rewrite never moves the region itself, only which side the final symbol points to.
Hunting for the optimum inside the region. For a nonconstant linear objective on a bounded region with solid boundaries, the optimum is attained on the boundary, and at least one corner reaches it. On such a region, evaluate the corners rather than interior points. Random interior points waste effort and miss the best value.

Key takeaways

A single linear inequality in two variables has a half-plane of solutions. The boundary line $ax + by = c$ splits the plane, and every point on one side satisfies the inequality while every point on the other does not.
Graph it by drawing the boundary solid for $\le$ or $\ge$ and dashed for $<$ or $>$ , then testing one point off the line to decide which side to shade.
A system of inequalities has the intersection of the half-planes as its solution set, the feasible region. The region’s corners are the solutions of the boundary-line systems, found by the same substitution or elimination you use for equations.
Linear programming maximizes or minimizes a linear objective over a feasible region. By the corner point principle, when an optimum exists and the region is a bounded polygon with corners, at least one vertex attains it. In that case, evaluate the objective at every corner and take the best.
A feasible region can be empty (contradictory constraints, nothing to optimize) or unbounded (an optimum may exist in one direction only). When an optimum exists and the region has corners, at least one corner attains it.

Systems of Inequalities

Learning goals

A single inequality carves the plane in two

Stacking inequalities: the feasible region

Optimizing over the region: linear programming

When the region is empty or unbounded

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)